Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Answer:
d) 0.1202 M
Explanation:
Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.
NaOH + HA → NaA + H₂O
The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:
22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol
The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.
The molar concentration of HA is:
2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M
